3
$\begingroup$

For a coherent state $|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum\frac{\alpha^n}{\sqrt{n!}}|n\rangle$, please show me how to prove,
$$ \mathrm{Tr}\left[|\alpha\rangle\langle\alpha|\hat{A}\right]=\langle\alpha|\hat{A}|\alpha\rangle, $$

where $\hat{A}$ is a quantum mechanics operator.

$\endgroup$
  • 1
    $\begingroup$ Use the fact that every unit vector can be completed to a Hilbert basis, then compute that trace with respect to the obtained basis since it is invariant under change of Hilbert basis. $\endgroup$ – Valter Moretti Oct 1 '15 at 10:43
  • $\begingroup$ This can also be seen as the cyclicity of the trace, $\operatorname{Tr}(BC)=\operatorname{Tr}(CB)$, with $B=|\alpha⟩:\mathbb C\to\mathcal H$ and $C=⟨\alpha|A:\mathcal H\to\mathbb C$. (For slightly more technical details see here.) The proof, however, is basis-based and along the lines of Gennaro's answer. $\endgroup$ – Emilio Pisanty Oct 1 '15 at 11:35
  • $\begingroup$ (And yes, I did write that comment exclusively so I could write 'cyclicity' and 'basis-based'.) $\endgroup$ – Emilio Pisanty Oct 1 '15 at 11:36
  • 1
    $\begingroup$ I disagree that this should be closed. It is a perfectly natural question and (versions of) this identity are around in lots of places ready to confuse any unwary undergrads that might pass by. It is an overall asset to the site. $\endgroup$ – Emilio Pisanty Oct 1 '15 at 23:17
11
$\begingroup$

Let $|n'\rangle$ be a basis of the Hilbert space, then $$ \textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rangle\langle n'|\alpha\rangle = \langle\alpha|A\left(\sum_{n'}|n'\rangle \langle n'|\right)|\alpha\rangle=\langle\alpha|A|\alpha\rangle $$

$\endgroup$
  • $\begingroup$ This is the procedure for a number state, but $|\alpha\rangle$ is a coherent state in this case. $\endgroup$ – TBBT Oct 1 '15 at 9:58
  • 5
    $\begingroup$ @TBBT No. This procedure works well for a coherent state and, in fact, for any state. The set $\{|n\rangle\}$ just needs to be any orthonormal basis. $\endgroup$ – higgsss Oct 1 '15 at 10:26
  • 1
    $\begingroup$ @TBBT As the above comment already mentioned, $|n\rangle$ is not a number state, it is just any element of a basis of a separable Hilbert space. You can call it $|\phi\rangle$ and integrate over instead, preferably. $\endgroup$ – gented Oct 1 '15 at 10:31
2
$\begingroup$

Another way to see this is to observe that any state $|\psi⟩\in\mathcal H$ can be extended to an orthonormal basis of the Hilbert space, and in that basis the trace $\operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right)$ is exactly $⟨\psi|\hat A|\psi⟩$.

More explicitly, for any $|\psi⟩\in\mathcal H$ there exists a sequence $\renewcommand{\phi}{\varphi}\left\{|\phi_n⟩\right\}_n$ such that $⟨\phi_n|\phi_m⟩=\delta_{nm}$, $⟨\phi_n|\psi⟩=0$, and $$|\psi⟩⟨\psi|+\sum_n|\phi_n⟩⟨\phi_n|=1.$$ In this basis, then, \begin{align} \operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right) =⟨\psi|\psi⟩⟨\psi|\hat A|\psi⟩+\sum_n⟨\phi_n|\psi⟩⟨\psi|\hat A|\phi_n⟩ =⟨\psi|\hat A|\psi⟩. \end{align}

For a coherent state $|\psi⟩=|\alpha⟩$, this can be made even more explicit by setting the basis as a displaced number state basis sitting on top of the coherent state, i.e. $|\phi_n⟩=\hat D(\alpha)|n⟩$ for $n=1,2,3,\ldots$ and $|n⟩$ a number state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.